Page 45 - Soil and water contamination, 2nd edition

32 Soil and Water Contamination

Rate of reverse reaction: r 2 k 2 [C ] [D ] (2.39b)
Net rate of the reaction: r k [A ] [B ] k [C ] [D ] (2.39c)
n 1 2
-3
-1
where r = reaction rate [M L T ], k = the rate constant of the forward reaction, k = rate
1 2
constant of reverse reaction (dimension depends on the exponents). The overall reaction
order is defined as the sum of the exponents in the rate expression. For the rate in the
forward reaction (Equation 2.39a), the reaction order is thus equal to α + β. The exponents
and the reaction order need to be derived experimentally.
In the absence of experimental evidence or other knowledge, the reactions in most
environmental models are often assumed to follow first-order kinetics , i.e. the reaction rate
is proportional to the activity or concentration of the reactant to the first power:

d [A ] 0 . 1
k [A ] k [A ] (2.40)
dt

-1
where k = the first-order reaction rate constant [T ]. Note that this kinetic equation
represents an irreversible process and can be applied in cases in which the reverse reaction
does not have a major effect, i.e. for a reaction of the form:
A products (2.41)

The analytical solution of this first-order differential Equation (2.40) reads:

A( t) A 0 e kt (2.42)

where A = the initial concentration (activity) of A at t = 0. The negative sign in front of the
0
rate constant indicates that the chemical is being removed, i.e. the concentration decreases
with time and goes to zero in the limit.
Example 2.8 Reaction order

Consider the decomposition of N O :
2 5
2N O → 4NO + O
2 5 2 2
The rate law for this reaction is
rate = -k [N O ]
2 5
Thus, the reaction is first-order with respect to the N O concentration. If the first-
2
5
-1
order reaction rate constant k = 0.062 s , calculate the time needed to lower the initial
concentration by 90 percent.
Solution
From Equation (2.42), we see that the N O concentration decreases exponentially over
2 5
time:
[N O ](t) = [N O ](t = 0) e -kt
2 5 2 5
After a certain time, the concentration has decreased to 10 percent of its initial value,
thus:
-kt
[N O ](t)/[N O ](t=0) = 0.1 = e = e -0.062 t
2 5 2 5

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